Equilibrium — AP Chemistry
1. What is Dynamic Chemical Equilibrium? ★★☆☆☆ ⏱ 3 min
Chemical equilibrium occurs for reversible reactions, denoted with the $\rightleftharpoons$ symbol. It is a dynamic, not static, state: the rate of the forward reaction equals the rate of the reverse reaction, so concentrations of reactants and products remain constant over time, with no net change to macroscopic properties like color, pressure, or pH. Equilibrium makes up 7-9% of your total AP Chemistry exam score, and applies to gas-phase, acid-base, and solubility systems.
2. Equilibrium Constants $K_c$ and $K_p$ ★★★☆☆ ⏱ 8 min
The equilibrium constant $K$ quantifies the ratio of products to reactants at equilibrium for a given reaction and temperature. Only temperature changes the value of $K$. For a general reversible reaction:
aA + bB \rightleftharpoons cC + dD
$K_c$ uses molar equilibrium concentrations for aqueous and gaseous species, written as:
K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}
Solids (s) and pure liquids (l) are always excluded from equilibrium expressions, as their chemical activity is 1, so they do not affect the ratio. Only aqueous (aq) and gaseous (g) species are included. For gas-phase reactions, $K_p$ uses partial pressures of gases instead of concentrations:
K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}
You can convert between $K_p$ and $K_c$ with the relationship:
K_p = K_c(RT)^{\Delta n}
Where $\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}$, $R = 0.0821 \text{ L·atm/(mol·K)}$, and $T$ is absolute temperature in Kelvin.
Exam tip: Examiners frequently test the rule excluding solids and pure liquids. Always cross out all non-aqueous, non-gaseous species before writing your equilibrium expression.
3. Reaction Quotient $Q$ ★★☆☆☆ ⏱ 4 min
The reaction quotient $Q$ uses the exact same formula as $K_c$ or $K_p$, but uses current non-equilibrium concentrations or partial pressures instead of equilibrium values. Comparing $Q$ to $K$ tells you which direction the reaction will shift to reach equilibrium:
- $Q < K$: The ratio of products to reactants is lower than equilibrium, so the reaction shifts right to make more products
- $Q = K$: The system is already at equilibrium, no net change occurs
- $Q > K$: The ratio of products to reactants is higher than equilibrium, so the reaction shifts left to make more reactants
4. Le Chatelier's Principle ★★★☆☆ ⏱ 6 min
Le Chatelier's principle states that if a stress (change in conditions) is applied to a system at equilibrium, the system will shift to counteract the stress and re-establish equilibrium. The three most common stresses are:
- **Concentration change**: Adding a reactant shifts equilibrium right to consume the added reactant; adding a product shifts left. Removing a species shifts equilibrium toward the side of the removed species.
- **Pressure/volume change**: Only affects gas-phase reactions with unequal moles of gas on each side. Increasing pressure (decreasing volume) shifts to the side with fewer moles of gas; decreasing pressure shifts to the side with more moles. If $\Delta n = 0$, pressure changes have no effect.
- **Temperature change**: The only stress that changes the value of $K$. For endothermic reactions ($\Delta H > 0$, heat is a reactant), increasing temperature shifts right and increases $K$. For exothermic reactions ($\Delta H < 0$, heat is a product), increasing temperature shifts left and decreases $K$.
Catalysts do not affect equilibrium position or $K$: they speed up forward and reverse reactions equally, only reducing the time to reach equilibrium.
5. Acid-Base Equilibria and Buffers ★★★☆☆ ⏱ 7 min
General equilibrium rules apply to weak acid and base dissociation in water. Start with the autoionization of water:
H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)
At 25°C, the ion product of water is $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$, and $pH + pOH = 14$, where $pH = -\log[H^+]$. For a weak acid $HA$ dissociating in water, the acid dissociation constant is:
K_a = \frac{[H^+][A^-]}{[HA]}
And $pK_a = -\log K_a$: smaller $pK_a$ means a stronger weak acid. Buffers are solutions that resist pH change when small amounts of strong acid or base are added. They are made of a weak acid and its conjugate base (or weak base and conjugate acid) in comparable concentrations. The Henderson-Hasselbalch equation calculates buffer pH:
pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)
Buffer capacity is highest when $pH = pK_a$ (equal concentrations of buffer components) and when total buffer concentration is high.
6. Solubility Product $K_{sp}$ ★★★★☆ ⏱ 6 min
The solubility product constant $K_{sp}$ describes the equilibrium between a sparingly soluble ionic solid and its dissolved ions in solution. For the general dissociation:
M_xA_y(s) \rightleftharpoons xM^{y+}(aq) + yA^{x-}(aq)
The solid is excluded from the $K_{sp}$ expression (as it is a pure solid), so:
K_{sp} = [M^{y+}]^x [A^{x-}]^y
Molar solubility $s$ is the moles of solid that dissolve per liter of solution, and can be related to $K_{sp}$ via dissociation stoichiometry. The common ion effect states that adding a soluble salt with an ion common to the sparingly soluble solid reduces its solubility.
Common Pitfalls
Why: You forget condensed phases have an activity of 1 and do not affect the equilibrium ratio
Why: You memorize pressure shift rules without checking the number of gas moles on each side
Why: You confuse shifts caused by Q changes with shifts caused by K changes
Why: You assume any acid-base mixture is a buffer
Why: You rush through problems and ignore the dissociation stoichiometry of the solid